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Question about schrodinger equation

  1. Dec 11, 2006 #1
    Note that this is not homework... im just curious

    The time independant Schrodinger equation can be written as
    [tex] \hat{H} \psi = E\psi[/tex]


    IS there ever a time taht the above equation is not true??

    what about the time dependant case?? We havent gone over that in class so im not quite sure about the cases.

    Thanks for your input!
     
  2. jcsd
  3. Dec 11, 2006 #2

    vanesch

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    The time-independent SE gives you those specific states that are stationnary. If you're looking for those states, then that equation is for you :smile:
    It is not a general solution to a quantum problem, but looks as specific kinds of cases. However, it also turns out that the solutions of this equation (which is simply the eigen-equation of the hamiltonian in fact) allows you to find the more general time-dependent solutions.

    As an analogy, consider it something like the static equilibrium equation F = 0 in Newtonian physics. It is not the general dynamical solution dp/dt = F, but only gives you the solutions of static equilibrium. However, here, it is less clear how the static solutions can help you find the general dynamic solutions, while in the quantum case, there is a clear link.
     
  4. Dec 11, 2006 #3
    As vanesch says, it only satisfies the stationary states. What it does is gives you the states of definite energy (which are useful for several reasons), in this case the energy is E.

    Compare with

    [tex]\hat{p}\psi=p\psi[/tex]

    which works for states of definite momentum, specifically of momentum p.

    Both of these are the eigenvalue equations for the respective operators. The time dependent equation (the Schrodinger equation proper) is given by

    [tex]\hat{H}| \psi \rangle = i\hbar \frac{d}{dt} | \psi \rangle[/tex]

    and this holds for any quantum state. One can almost (problems arise with spin-statistics) view this equation as defining which [itex]| \psi \rangle[/itex] are allowable quantum states. In the case of time-independent Hamiltonians, we can separate the time dependence out, as I'm sure you will soon find out in class.
     
  5. Dec 11, 2006 #4
    The answers above are great, but here are some shorter ones.

    Yes.

    The above equation does not hold for time-dependent cases. i.e. when the wave function changes shape over time. A simple example is a particle moving through empty space (as a plane wave).

    To make the time dependent form we replace the constant E with an expression which can handle the variations over time [itex]\mathrm{i} \hbar \frac{\partial \psi}{\partial t}[/itex]:

    [tex] \hat{H} \psi = \mathrm{i} \hbar \frac{\partial \psi}{\partial t}[/tex]

    AFAIK, this is identical to the form given in the previous posts with kets.

    Hope this helps.
     
  6. Dec 11, 2006 #5
    so stationary states are defined at hte square modulus of the wave function not having a time dependance.
    How would we know this before solving for the wavefunction? If it was time independant then certain [itex] \psi [/itex] would not depend on time

    would this mean taht all solutions of the TISE (time indepdnant schrodinger equation) are stationary states?? Since the solution for the time dependance is always [itex] \exp(-iEt/\hbar) [/itex]. If we could do that then certainly our potential is not time varying. Hence there's never a time depdance and the state is stationary always?
     
  7. Dec 11, 2006 #6
    We don't, but if we do know then we can use that to simplify the problem. If we try and solve a dynamic problem with the time-independent equation then we should eventually end up with some nonsensical expression.

    It should be quite obvious when to use which one. Since ground states shouldn't change spontaneously we can use the time-independent equation to approximate them. If we wish to model an evolving state then we need to use the time-dependent form.


    I think I may have made a mistake in the formula somewhere though. I keep thinking that [itex]\frac{\partial \psi}{\partial t}[/itex] is equal to zero when nothing is changing over time, so that we should have [itex]\hat{H} \psi = 0[/itex] for the time-independent form. So, I'm not sure exactly why the E's are not zero... probably some oversight on my part...
     
  8. Dec 11, 2006 #7
    Time-independence of a physical system, only requires the square modulus of the wavefunction to be time-independent, not the wave function itself. This means that the wave function itself can vary with time, although only in a way that does not change its sqaure modulus, which means that it can only change by phasing, in which case [itex]\hat{H} \psi = E \psi[/itex]
     
  9. Dec 11, 2006 #8
    Ah, of course, not sure why I forgot that, it is even mentioned already in this thread. Its also pretty damned important for calculations... oh well.
     
  10. Dec 11, 2006 #9
    change the phasing...

    where is the pahse factor appearing in the wave equation?
    Lets say for the wavefunction of hte infinite square well

    [tex] \Psi(x,t) = N \sin\frac{n\pi x}{a} e^{-\frac{\i E_{n}t}{\hbar}} [/tex]

    well a wave can also be written like
    [tex] Asin(kx-\phi) + Bcos(kx-\phi) [/tex]

    i dont quite see how the phi factor comes about in the wave equation... where would it be??
    could it be like this?
    [tex] \Psi(x,t) = N \sin\left(\frac{n\pi x}{a}+\Phi\right) e^{-\frac{\i E_{n}t}{\hbar}} [/tex]
     
  11. Dec 11, 2006 #10

    jtbell

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    You can add an arbitrary constant complex phase factor to a QM wave function, without changing the physics:

    [tex] \Psi(x,t) = N \sin\left(\frac{n\pi x}{a} \right) e^{-i \left( \frac{ E_{n}t}{\hbar} + \phi \right)} [/tex]


    Physically, this corresponds to the fact that the "zero point" of potential energy (and therefore also of total energy) is arbitrary. Note what happens when you set [itex]V(x) = V_0[/itex] (a constant) inside the infinite square well, and then solve the Schrödinger equation again.
     
    Last edited: Dec 11, 2006
  12. Dec 11, 2006 #11
    then wouldnt we have to consider different situations for the solution suppose E> V0, E<V0, E=V0??

    three possible solutions then right? Just wanna make sure before i go ahead with this. The wavefunction would still ahve to vanish at the boundary points, though.
     
  13. Dec 11, 2006 #12

    vanesch

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    Except that it won't give you a constant phase factor, but one of the form [tex]e^{-i V_0 t/\hbar}[/tex]
     
  14. Dec 11, 2006 #13

    jtbell

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    Oops, you're right. :blushing:

    I should have realized that "relocating" the bottom of the well would simply change [itex]E_n[/itex] to [itex]E_n - V_0[/itex].

    The arbitrary [itex]\phi[/itex] simply indicates the "initial value" of the complex oscillation, and doesn't correspond to anything physical. Any time you calculate an expectation value or probability, it always contributes a net factor of [itex]e^{i \phi} e^{-i \phi} = 1[/itex].

    This is true only for a [itex]\phi[/itex] that doesn't depend on x, of course. If [itex]\phi[/itex] depends on x, that is, the initial phase is different at different locations instead of being uniform everywhere, things get interesting! :biggrin:
     
  15. Dec 11, 2006 #14

    jtbell

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    In principle yes, but you already have to do that when [itex]V_0 = 0[/itex], i.e. consider the cases E > 0, E < 0, E = 0.

    For [itex]E < V_0[/itex] you'll find that it's impossible to construct a wave function that vanishes at both sides of the well. Have you done the finite square well ([itex]V = V_0[/itex] outside the well, and 0 inside) yet, or tunneling through a barrier (V = 0 outside and [itex]V = V_0[/itex] inside)? These are the situations where people usually first tackle [itex]E < V_0[/itex].
     
  16. Dec 12, 2006 #15
    yes i have done the tunnelling cases
    so if its impossible to make the wavefunction vanish at the end points then what boundary conditions could we apply??
     
  17. Dec 12, 2006 #16

    jtbell

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    We don't have any choice in the boundary conditions. The boundary conditions are fixed by the choice of potential. For an infinite square well, we must require that [itex]\psi = 0[/itex] at the boundaries. If a general mathematical solution of the S.E. with a particular E cannot satisfy that condition, then it is not admissible as a physical solution for this situation.

    As you've seen in the tunnelling case, the general solution for E < V involves real exponentials: [itex]\psi = Ae^{kx} + Be^{-kx}[/itex]. There's no way to set the arbitrary constants A and B so that [itex]\psi = 0[/itex] at both boundaries of the infinite square well. Therefore it is impossible to have E < V for that situation.
     
  18. Dec 13, 2006 #17
    Oh i have another question about this
    Is this always true??
    [tex] \hat{H} \Psi = E \Psi [/tex]
    Here we talk about the wavefunction in general
    I think this is not always true ... but i dont know understand why?? I m thinking that it ahs something to do with the time dependant case where the right is
    [tex] i\hbar \frac{d}{dt} \Psi [/tex]
    this term need not repsent the energy... right?
     
  19. Dec 14, 2006 #18

    dextercioby

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    You've already been explained the difference between a time-dependent Hamiltonian in Schroedinger picture and a time-independent one. And how that determins the time-evolution of the state vector.

    [tex] \hat{H} \Psi = E \Psi [/tex]

    is a spectral equation for an operator on separable Hilbert space and nothing more. It has solutions in the Hilbert space iff the spectrum of the Hamiltonian is discreet.

    Daniel.
     
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